kelvind13 wrote:

I too arrived at the same solution, although it took me close to about 4 minutes to do it. isn't there a faster shortcut for the same?

Bunuel wrote:

PrepTap wrote:

Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

A. 1/6

B. 1/16

C. 1/18

D. 1/36

E. 25/216

Total outcomes:6^3.

Favorable outcomes:{6 - 5 - 1} - 6 cases ({6 - 5 - 1}, {6 - 1 - 5}, {5 - 6 - 1}, {5 - 1 - 6}, {1 - 6 - 5}, {1 - 5 - 6});

{6 - 4 - 2} - 6 cases;

{6 - 3 - 3} - 3 cases;

{5 - 5 - 2} - 3 cases;

{5 - 4 - 3} - 6 cases;

{4 - 4 - 4} - 1 case.

6 + 6 + 3 + 3 + 6 + 1 = 25.

P = (favorable)/(total) = 25/6^3.

Answer: E.

As such there is no shortcut possible but the counting can be done in an orderly fashion which might reduce the time you took.

We know that the sum required is 12.

So if in the first place we have:

1 -> Sum of other two numbers = 12 -1 = 11 -> possible combinations (5+6, 6+5)

2 -> Sum of other two numbers = 12 -2 = 10 -> possible combinations (4+6, 6+4, 5+5)

3 -> Sum of other two numbers = 12 -3 = 9 -> possible combinations (3+6, 6+3, 4+5, 5+4)

4 -> Sum of other two numbers = 12 - 4 = 8 -> possible combinations (2+6, 6+2, 3+5, 5+3, 4+4)

5 -> Sum of other two numbers = 12 -5 = 7 -> possible combinations (1+6, 6+1, 2+5, 5+2, 3+4, 4+3)

6 -> Sum of other two numbers = 12 -6 = 6 -> possible combinations (1+5, 5+1, 2+4, 4+2, 3+3)

Favourable outcomes: 25

Total outcomes: 6 * 6 * 6 = 216

Probability = 25/216